Show that the area enclosed between the curve `y^(2)=12(x+3) and y^(2)=20(5-x)` is `64sqrt((5)/(3))`

Answer

To find the area enclosed between the curves given by the equations \( y^2 = 12(x + 3) \) and \( y^2 = 20(5 - x) \), we will follow these steps:

### Step 1: Find the Points of Intersection
We start by equating the two equations since both are equal to \( y^2 \):

\[ 12(x + 3) = 20(5 - x) \]

Expanding both sides:

\[ 12x + 36 = 100 - 20x \]

Rearranging the equation gives:

\[ 12x + 20x = 100 - 36 \]

\[ 32x = 64 \]

Dividing both sides by 32:

\[ x = 2 \]

Now, substitute \( x = 2 \) back into either equation to find \( y \):

Using the first equation:

\[ y^2 = 12(2 + 3) = 12 \times 5 = 60 \]

Thus, \( y = \pm 2\sqrt{15} \).

The points of intersection are \( (2, 2\sqrt{15}) \) and \( (2, -2\sqrt{15}) \).

### Step 2: Determine the Limits of Integration
The curves intersect at \( x = 2 \). We need to find the x-coordinates where the curves intersect the x-axis.

From the first equation \( y^2 = 12(x + 3) \):

Setting \( y = 0 \):

\[ 0 = 12(x + 3) \implies x + 3 = 0 \implies x = -3 \]

From the second equation \( y^2 = 20(5 - x) \):

Setting \( y = 0 \):

\[ 0 = 20(5 - x) \implies 5 - x = 0 \implies x = 5 \]

Thus, the limits of integration are from \( x = -3 \) to \( x = 5 \).

### Step 3: Set Up the Integral for Area
The area \( A \) between the curves can be computed using the formula:

\[ A = \int_{-3}^{2} (y_{\text{upper}} - y_{\text{lower}}) \, dx + \int_{2}^{5} (y_{\text{upper}} - y_{\text{lower}}) \, dx \]

Here, \( y_{\text{upper}} = 2\sqrt{5 - x} \) and \( y_{\text{lower}} = -2\sqrt{3 + x} \).

### Step 4: Calculate the Area
We can express the area as:

\[ A = 2 \left( \int_{-3}^{2} 2\sqrt{3 + x} \, dx + \int_{2}^{5} 2\sqrt{5 - x} \, dx \right) \]

Calculating the first integral:

\[ \int 2\sqrt{3 + x} \, dx \]

Using the substitution \( u = 3 + x \), \( du = dx \):

\[ = 2 \cdot \frac{2}{3} (3 + x)^{3/2} \Big|_{-3}^{2} = \frac{4}{3} \left( (5)^{3/2} - (0)^{3/2} \right) = \frac{4}{3} \cdot 5\sqrt{5} \]

Calculating the second integral:

\[ \int 2\sqrt{5 - x} \, dx \]

Using the substitution \( v = 5 - x \), \( dv = -dx \):

\[ = -2 \cdot \frac{2}{3} (5 - x)^{3/2} \Big|_{2}^{0} = \frac{4}{3} \left( (5)^{3/2} - (0)^{3/2} \right) = \frac{4}{3} \cdot 5\sqrt{5} \]

### Final Calculation
Combining both areas:

\[ A = 2 \left( \frac{4}{3} \cdot 5\sqrt{5} + \frac{4}{3} \cdot 5\sqrt{5} \right) = 2 \cdot \frac{8}{3} \cdot 5\sqrt{5} = \frac{80\sqrt{5}}{3} \]

### Conclusion
Thus, the area enclosed between the curves is:

\[ A = 64\sqrt{\frac{5}{3}} \text{ square units.} \]